Question

Initially a particle is moving at 4.40m/s at an angle of 37.5 degree above the horizontal. Two seconds later, it's velocity is 6.25m/s at an angle of 56.0 degrees below the horizontal. What was the particles average acceleration during these 2.00 seconds?

Answer 1

Answer:

3.93 m/s² in the south direction.

Explanation:

We need to find the velocities in each direction.

So, initially, the particle is moving at 4.40 m/s at an angle of 37.5 degrees. Then, this velocity has a horizontal and vertical component equal to:

$\begin{gathered} v_{ix}=v_i\cos (37.5)=4.40\cos (37.5)=3.49 \\ v_{iy}=v_i\sin (37.5)=4.40\sin (37.5)=2.68 \end{gathered}$

In the same way, in the end, the particle is moving at 6.25 m/s at an angle of -56.0 degrees, so each component is equal to:

$\begin{gathered} v_{fx}=v_f\cos (-56)=6.25\cos (-56)=3.49 \\ v_{fy}=v_f\sin (-56)=6.25\sin (-56)=-5.18 \end{gathered}$

Now, we can calculate the average acceleration on each component using the following equation:

$\begin{gathered} a_x=\frac{v_{fx}-v_{ix}}{t}=\frac{3.49-3.49}{2}=0m/s^2 \\ a_y=\frac{v_{fy}-v_{iy}}{t}=\frac{-5.18-2.68}{2}=-3.93m/s^2_{} \end{gathered}$

Therefore, the average acceleration is 3.93 m/s² in the south direction.